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RE: Math problem (triangle area solution)

To: "Matt Murray" <mattm@optonline.net>, <general@rennlist.org>,
Subject: RE: Math problem (triangle area solution)
From: "Adamson, Ken" <KADAMSON@lifeline.net>
Date: Wed, 5 Nov 2003 17:57:23 -0600
Ok, I can remember all this stupid trig, but I can't multiply two numbers 
together...

Last try:

Law of cosines

c2 = a2+b2-2abCosC

190*190 = 125*125+170*170 - 2*125*170*CosC      (substitute)

36100 = 15625 + 28900 - (42500* CosC)           (do some math)

-8425 = -(42500 * CosC)                                 (do some algebra)

8425 = 42500 * CosC                                     (some more algebra)

CosC = 8425/42500                                       (take the inverse 
cosine)

C = 78.566 degress

Now we have an included angle, so we can switch to the area formula:

K = 1/2 * abSinC

K = .5 * 125 * 170 * Sin(78.566)

K = 10414.141 square units

That seems better :)


Ken
'01 PT Cruiser Limited
HS101 in OKC (soon to be STS101)





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